Re: [R] p-values from coxph?

2010-10-18 Thread Terry Therneau 
On Oct 15, 2010, at 9:21 AM, ?hagen Patrik wrote:


 Dear List,

 I each iteration of a simulation study, I would like to save the p- 
 value generated by coxph. I fail to see how to adress the p-value.  
 Do I have to calculate it myself from the Wald Test statistic?

No.  Look at help(coxph.object).  This list the components of a coxph
object and explains what they are.  You will find that
   fit - coxph(.
   fit$wald.test

contains the Wald test statistic.  I prefer the likelihood ratio test
myself 2*diff(fit$loglik), with fit$df degrees of freedom.

Hunting with str(...) is a good strategy, but even better is the
documentation (when it exists).

Terry T.

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[R] p-values from coxph?

2010-10-15 Thread Öhagen Patrik 

Dear List,

I each iteration of a simulation study, I would like to save the p-value 
generated by coxph. I fail to see how to adress the p-value. Do I have to 
calculate it myself from the Wald Test statistic?


Cheers, Paddy

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Re: [R] p-values from coxph?

2010-10-15 Thread David Winsemius 


On Oct 15, 2010, at 9:21 AM, Öhagen Patrik wrote:



Dear List,

I each iteration of a simulation study, I would like to save the p- 
value generated by coxph. I fail to see how to adress the p-value.  
Do I have to calculate it myself from the Wald Test statistic?


No. And the most important reason is that would not give you the same  
value as is print()-ed by coxph().


If you ask for the the str(print(coxph(...)) you get NULL (after the  
side-effect of prinitng. The print function only produces side- 
effects. On the other hand you can use the summary function and it  
gives you a richer set of output.  Using the first example on the help  
page for coxph:


str(summary(coxph(Surv(time, status) ~ x + strata(sex), test1)))
List of 12
 $ call: language coxph(formula = Surv(time, status) ~ x +  
strata(sex), data = test1)

 $ fail: NULL
 $ na.action   : NULL
 $ n   : int 7
 $ loglik  : num [1:2] -3.87 -3.33
 $ coefficients: num [1, 1:5] 0.802 2.231 0.822 0.976 0.329
  ..- attr(*, dimnames)=List of 2
  .. ..$ : chr x
  .. ..$ : chr [1:5] coef exp(coef) se(coef) z ...
 $ conf.int: num [1, 1:4] 2.231 0.448 0.445 11.18
  ..- attr(*, dimnames)=List of 2
  .. ..$ : chr x
  .. ..$ : chr [1:4] exp(coef) exp(-coef) lower .95 upper .95
 $ logtest : Named num [1:3] 1.087 1 0.297
  ..- attr(*, names)= chr [1:3] test df pvalue
 $ sctest  : Named num [1:3] 1.051 1 0.305
  ..- attr(*, names)= chr [1:3] test df pvalue
 $ rsq : Named num [1:2] 0.144 0.669
  ..- attr(*, names)= chr [1:2] rsq maxrsq
 $ waldtest: Named num [1:3] 0.95 1 0.329
  ..- attr(*, names)= chr [1:3] test df pvalue
 $ used.robust : logi FALSE

So the fifth element of  coefficients leaf of the list structure has  
the same p-value as that print()-ed.


Try:

 summary(fit)$coefficients[5]
[1] 0.3292583

(It does seem to me that the name for that leaf of the fit object is  
not particularly in accord with what I would have considered  
coefficients., but I am really in no solid position to criticize  
Terry Therneau to whom we all owe a great deal of gratitude.)



--
David.
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